6)2xy+x=6y+8
7)xy+12=x+y
8)x^2+xy-x-y=0
9)xy^2-xy+y-3=0
10)xy+7x-2y-14=0
Tìm số nguyên x biết
a,3x+3y-2xy=7
b,xy+2x+y+11=0
c,xy+x-y=4
d,2x.(3y-2)+(3y-2)=12
e,3x+4y-xy=15
f,xy+3x-2y=11
g,xy+12=x+y
h,xy-2x-y=-6
i,xy+4x=25+5y
ii,2xy-6y+x=9
iii,xy-x+2y=3
k,2.x^2.y-x^2-2y-2=0
l,x^2.y-x+xy=6
Phân tích đa thức 3\(x^2\)y + 6\(xy^2\) – 9xy thành nhân tử. Kết quả là:
A. 3(\(x^2y\) + 2\(xy^2\) – 3xy - 3). B. 3y(\(x^2\) + 2xy – 3x). C. xy(3x + 6y - 9). D. 3xy(x + 2y – 3).
Giải hpt
\(\left\{\begin{matrix}xy^2 -3xy + 3x -2y +2 = 0\\ x^2 + y^2 +xy-7x-6y+14= 0\\ \end{matrix}\right.\)
\(\hept{\begin{cases}xy^2-3xy+3x-2y+2=0\\x^2+y^2+xy-7x-6y+14=0\end{cases}}\)
HPT \(\Leftrightarrow\hept{\begin{cases}x\left(y^2-4y+4\right)+xy-x-2y+2=0\\\left(x^2-4x+4\right)+\left(y^2-4y+4\right)+xy-2x-2y+4-x+2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\left(y-2\right)^2+\left(x-2\right)\left(y-2\right)+\left(x-2\right)=0\\\left(x-2\right)^2+\left(y-2\right)^2+\left(x-2\right)\left(y-2\right)-\left(x-2\right)=0\end{cases}}\)
Đặt a = x - 2 ; b = y - 2 ta có :
\(\hept{\begin{cases}\left(a+2\right)b^2+ab+a=0\\a^2+b^2+ab-a=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a\left(b^2+b+1\right)=-2b^2\\a=a^2+b^2+ab\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a=\frac{-2b^2}{b^2+b+1}\le0\forall b\\a=a^2+b^2+ab\ge0\forall ab\end{cases}}\)
\(\Rightarrow a=0\Rightarrow b=0\Rightarrow x=y=2\left(TM\right)\)
1)x=6y và |x|-|y|=60
2) |x| +|y| <2
3) (x+1)^2 +(y+1)^2 +(x-y)^2 =2
4) (x-2)(5y+1)=12
5) (8– x)(4y +1) = 20
6) xy = x+y
7) x(y+2)+y =1
8) (x-2)(xy-1)=5
9) (2x+1)(y- 5)=12
10) (x-4)(2y+1)=7
11) (2x +1)(3y – 2) = -33
12) xy +5x- 7y= 35
13) xy +2x-3y= 9
14) xy-2x+5y-12=0
1.
a.(-xy)(-2x2y+3xy-7x)
b.(1/6x2y2)(-0,3x2y-0,4xy+1)
c.(x+y)(x2+2xy+y2)
d.(x-y)(x2-2xy+y2)
2.
a.(x-y)(x2+xy+y2)
b.(x+y)(x2-xy+y2)
c.(4x-1)(6y+1)-3x(8y+4/3)
1.
\(a,\left(-xy\right)\left(-2x^2y+3xy-7x\right)\)
\(=2x^3y^2-3x^2y^2+7x^2y\)
\(b,\left(\dfrac{1}{6}x^2y^2\right)\left(-0,3x^2y-0,4xy+1\right)\)
\(=-\dfrac{1}{20}x^4y^3-\dfrac{1}{15}x^3y^3+\dfrac{1}{6}x^2y^2\)
\(c,\left(x+y\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x+y\right)^3\)
\(=x^3+3x^2y+3xy^2+y^3\)
\(d,\left(x-y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x-y\right)^3\)
\(=x^3-3x^2y+3xy^2-y^3\)
2.
\(a,\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3-y^3\)
\(b,\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=x^3+y^3\)
\(c,\left(4x-1\right)\left(6y+1\right)-3x\left(8y+\dfrac{4}{3}\right)\)
\(=24xy+4x-6y-1-24xy-4x\)
\(=\left(24xy-24xy\right)+\left(4x-4x\right)-6y-1\)
\(=-6y-1\)
#Toru
1)\(\begin{cases}x^4+2xy+6y-7x^2-2x^2y+9=0\\2x^2y-x^3=10\end{cases}\)
2)\(\begin{cases}2x^3-x^2y+x^2+y^2-2xy-y=0\\xy+x-2=0\end{cases}\)
a)x^2(x-3)-4x+12 b)2a(x+y)-x+y c)6x^2-12x-7x+14 d)xy-y^2-3x+3y f)x^2y+xy^2-4x-4y g)10ax-5ay-7x+14 j)a^3-a^2+9a-9(tính nhân tử chung)
a: \(x^2\left(x-3\right)-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
b: \(2a\left(x+y\right)+x+y=\left(x+y\right)\left(2a+1\right)\)
c: \(6x^2-12x-7x+14\)
\(=6x\left(x-2\right)-7\left(x-2\right)\)
\(=\left(x-2\right)\left(6x-7\right)\)
Tìm x,y thuộc Z,biết : a) xy+5x+y=4 b)xy+14+2y+7x=-10 c)xy+x+y=2.
`a)xy+5x+y=4`
`=>x(y+5)+y+5=9`
`=>(y+5)(x+1)=9`
Vì `x,y in ZZ`
`=>x+1,y+5 in ZZ`
`=>x+1,y+5 in Ư(9)={+-1,+-3,+-9}`
Đến đây xét giá trị rồi giải(cái này phải tự làm).
`b)xy+14+2y+7x=0`
`=>y(x+2)+7(x+2)=0`
`=>(x+2)(y+7)=0`
`=>` \(\left[ \begin{array}{l}x=-2\\y=-7\end{array} \right.\)
`c)xy+x+y=2`
`=>x(y+1)+y+1=3`
`=>(x+1)(y+1)=3`
Vì `x,y in ZZ`
`=>x+1,y+1 in ZZ`
`=>x+1,y+1 in Ư(3)={+-1,+-3}`
Đến đây xét giá trị rồi giải(cái này phải tự làm).
Giải:
a) \(xy+5x+y=4\)
\(\Rightarrow x.\left(y+5\right)+\left(y+5\right)=9\)
\(\Rightarrow\left(x+1\right).\left(y+5\right)=9\)
\(\Rightarrow\left(x+1\right)\) và \(\left(y+5\right)\inƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)
Ta có bảng giá trị:
x+1 | -9 | -3 | -1 | 1 | 3 | 9 |
y+5 | -1 | -3 | -9 | 9 | 3 | 1 |
x | -10 | -4 | -2 | 0 | 2 | 8 |
y | -6 | -8 | -14 | 4 | -2 | -4 |
Vậy \(\left(x;y\right)=\left\{\left(-10;-6\right);\left(-4;8\right);\left(-2;-14\right);\left(0;4\right);\left(2;-2\right);\left(8;-4\right)\right\}\)
b) \(xy+14+2y+7x=-10\)
\(\Rightarrow y.\left(x+2\right)+7x+14=-10\)
\(\Rightarrow y.\left(x+2\right)+7.\left(x+2\right)=-10\)
\(\Rightarrow\left(x+2\right).\left(y+7\right)=-10\)
\(\Rightarrow\left(x+2\right)\) và \(\left(y+7\right)\inƯ\left(-10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
Ta có bảng giá trị:
x+2 | -10 | -5 | -2 | -1 | 1 | 2 | 5 | 10 |
y+7 | 1 | 2 | 5 | 10 | -10 | -5 | -2 | -1 |
x | -12 | -7 | -4 | -3 | -1 | 0 | 3 | 8 |
y | -6 | -5 | -2 | 3 | -17 | -12 | -9 | -8 |
Vậy \(\left(x;y\right)=\left\{\left(-12;-6\right);\left(-7;-5\right);\left(-4;-2\right);\left(-3;3\right);\left(-1;-17\right);\left(0;-12\right);\left(3;-9\right);\left(8;-8\right)\right\}\)
c) \(xy+x+y=2\)
\(\Rightarrow x.\left(y+1\right)+\left(y+1\right)=3\)
\(\Rightarrow\left(x+1\right).\left(y+1\right)=3\)
\(\Rightarrow\left(x+1\right)\) và \(\left(y+1\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
x+1 | -3 | -1 | 1 | 3 |
y+1 | -1 | -3 | 3 | 1 |
x | -4 | -2 | 0 | 2 |
y | -2 | -4 | 2 | 0 |
Vậy \(\left(x;y\right)=\left\{\left(-4;-2\right);\left(-2;-4\right);\left(0;2\right);\left(2;0\right)\right\}\)
Chúc bạn học tốt!
Tìm x, y,z ∈ sao cho :
1) x+y+xy = 0
2) xy-x-y = 0
3) xy+x+y = 0
4) xy+2x-2y = 5
5) xy+2x+2y= -1
6) xy-2x-2y = -3
7) 2xy-x+y = 3
8) 3xy+2x-y = 2 ( HD:nhân 2 vế với 3)
Giúp mình vs...Gấp lắm r
Mỗi lần làm là 1 lần tick nhé, cảm ơn trc